Understanding Significant Figures in Physics: Why Only 2 Sig. Figs. for t?

  • Thread starter e-zero
  • Start date
In summary, the conversation discusses the calculation of the time it took for King Kong to fall from the top of the Empire State Building, using the formula x = x0 + v0t + (1/2)at^2. The calculated time was 8.80631 seconds, rounded to 8.81 seconds. However, the book reports the answer as 8.8 seconds, which is only two significant figures. The reason for this discrepancy is the differing conventions on significant figures and the assumption of acceleration due to gravity. The conversation also mentions the use of scientific notation to remove ambiguity in significant figures and the fact that the standard acceleration of gravity varies in different cities.
  • #1
e-zero
58
0
I have the following info for a question:

x0 = 380m
a = -9.80m/s^2
t = ?
v0 = 0

I use the formula x = x0 +v0t + (1/2)at^2

When I solve for 't' I get 8.80631

I figure I would round to 3 sig. figs. and have t = 8.81s, but my book is telling me t = 8.8 which is only 2 sig. figs.

Why are they reporting only 2 sig. figs. for this question?
 
Physics news on Phys.org
  • #2
I agree 8.81s looks better. Apparently authors of the book were not treating significant figures too religiously. Please remember they are only an approximated way of dealing with precision, and not a very good one.
 
  • #3
Here is the full question in case I missed something:

Estimate how long it took King Kong to fall straight down from the top of the Empire State Building (380m high).

I assume a = -9.80m^2. The only other thing I see is that the height of the building can be seen as only 2 sig. figs. since the last digit is '0', but I would normally take that as 3 sig. figs. since it is not stated as 'approximate'.

Any input?
 
  • #4
You haven't missed anything.
 
  • #5
In questions involving the acceleration due to gravity at the Earth's surface the key word "Estimate" usually means "assume a = 10 ms-2". So an estimate would be the square root of 76, or 8.7.

8.81 would be more appropriate if the question said "Calculate..."
 
  • #6
There is an ambiguity when dealing with significant figures and different people/textbooks use different conventions.

Often 380m indicates two significant figures.
Including a decimal point after the last digit (380.m) would indicate three.

Using scientific notation removes the ambiguity
3.8 is 2 sig figs
3.80 is 3
3.8000 is 5

Many of the introductory physics books I've seen use this convention.
 
  • #7

Related to Understanding Significant Figures in Physics: Why Only 2 Sig. Figs. for t?

1. Why is it important to use only 2 significant figures in scientific measurements?

Using only 2 significant figures helps to maintain consistency and accuracy in scientific data. It eliminates unnecessary digits and allows for easier comparison and calculation of data.

2. Can using more than 2 significant figures in scientific measurements affect the accuracy of the data?

Yes, using more significant figures than necessary can lead to a false sense of precision and can introduce errors in calculations and comparisons. It is important to use only the significant figures that are known with certainty.

3. How do you determine which digits are significant in a measurement?

In general, all non-zero digits are considered significant. Zeros can also be significant, depending on their location in the number. Leading zeros and trailing zeros after a decimal point are significant, while trailing zeros before a decimal point may not be significant.

4. Are there any exceptions to the rule of using only 2 significant figures in scientific measurements?

Yes, there are some cases where more significant figures may be necessary, such as when using logarithmic or exponential functions, or when reporting extremely precise measurements. However, for most scientific calculations, 2 significant figures is sufficient.

5. How does using only 2 significant figures affect the uncertainty of a measurement?

Using only 2 significant figures can help to reduce the uncertainty of a measurement, as it allows for a more accurate representation of the data. However, the uncertainty of a measurement is also affected by other factors, such as the precision of the measuring instrument and the skill of the person taking the measurement.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
659
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
5K
  • Other Physics Topics
Replies
2
Views
950
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
632
  • Introductory Physics Homework Help
Replies
1
Views
900
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Classical Physics
Replies
0
Views
208
  • Biology and Chemistry Homework Help
Replies
7
Views
2K
Back
Top